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Math Help - two variabe induction question..

  1. #1
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    two variabe induction question..

    n1 is the smallest whole number for which this inequality works :

    (1+x)^n >1 +n*x + n*x^2

    also i am given that x>0

    find n1

    and prove this inequality for every n=>n1 by induction.

    the base case:

    (1+x)^n1 >1+n1*x + n1*x^2

    i think its correct because i was told that this inequality works for n1.

    n=k step we presume that this equation is true :

    equation 1: (1+x)^k >1 +k*x + k*x^2

    n=k+1 step we need to prove this equation:
    equation 2: (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

    now i need to multiply equation1 by sum thing
    and
    do
    if a<b<c
    then a<c

    how to do thing in this case?
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  2. #2
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    equation 1: (1+x)^k >1 +k*x + k*x^2
    n=k+1 step we need to prove this equation:
    equation 2: (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2


    Just multiply equation 1 by (1+x) which is > 0
    (1+x)^(k+1) >(1 +k*x + k*x^2) (1+x)
    Then develop the second member
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  3. #3
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    i tried to do this
    but its not working
    (1+x)^(k+1) >(1 +k*x + k*x^2) (1+x)>(1 +k*x + k*x^2)


    (1+x)^(k+1) >(1 +k*x + k*x^2)

    ??
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  4. #4
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    It is !
    (1 + kx + kx) (1+x) = 1 + (k+1)x + 2kx + kx^3 > 1 + (k+1)x + 2kx > 1 + (k+1)x + (k+1)x because 2k > k+1
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  5. #5
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    i need to prove that equation 2 is true:
    (1+x)^(k+1) >1 +(k+1)*x + (k+1)*x^2

    i take equation1 and multiply it by (1+x)

    (2+x)^k >(1 +k*x + k*x^2)(1+x)

    but i dont know how to involve equation2 in here

    ??
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    i take equation1 and multiply it by (1+x)

    (2+x)^k >(1 +k*x + k*x^2)(1+x)

    but i dont know how to involve equation2 in here

    ??
    Taking equation 1
    (1+x)^k >1 +k*x + k*x^2
    and multiplying by (1+x) gives
    (1+x)^k (1+x)>(1 +k*x + k*x^2)(1+x)
    (1+x)^(k+1)>1 + (k+1)x + 2kx + kx^3 > 1 + (k+1)x + 2kx > 1 + (k+1)x + (k+1)x because 2k > k+1

    Remember that (1+x)^k (1+x)=(1+x)^(k+1)
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  7. #7
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    why (1+x)^k (1+x)=(1+x)^(k+1)
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  8. #8
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    Because
    (1+x)^k------=(1+x)(1+x) ... (1+x)--------------> k times
    (1+x)^k (1+x)=(1+x)(1+x) ... (1+x)(1+x)---------> k+1 times

    Therefore (1+x)^k (1+x)=(1+x)^(k+1)
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  9. #9
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    aaahhh ok thanks


    i am being asked to find n1

    how to do that
    ??
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  10. #10
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    Well you can see that it does not work for n=1 and n=2 but it works for n=3
    I would say n1=3
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  11. #11
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    i can replace n1 with n

    (1+x)^n1 >1 +n1*x + n1*x^2

    (1+x)^n1>1+n1*(x+x^2)

    how to separate n1 here?

    is this the right way??
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  12. #12
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    I am afraid you cannot separate n1 in equation (1+x)^n1 >1 +n1*x + n1*x^2
    The only way I can see to find n1 is the one I gave you in my previous post
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  13. #13
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    i did that


    if n=1
    (1+x)>1+x+x^2

    if n=2

    (1+x)^2 >1 +2*x + 2*x^2

    if n=3

    (1+x)^3>1+3*x+3*x^2

    these cases dont work

    what to do?
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  14. #14
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    I do not understand how you calculate

    if n=1
    (1+x)<1+x+x^2 => does not work

    if n=2
    (1+x)^2 = 1+2x+x < 1 +2*x + 2*x^2 => does not work

    if n=3
    (1+x)^3 = 1+3x+3x+x^3 >1+3*x+3*x^2 => works
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  15. #15
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    thanks
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